First slide

Trigonometric transformations

Question

$\frac{\sin^{2} A - \sin^{2} B}{\sin Acos A - \sin Bcos B}$ is equal to

Moderate

A

tan(A - B)
B

tan(A + B)
C

cot(A - B)
D

cot(A + B)

Solution

$\begin{array}{l} \frac{\sin^{2} A - \sin^{2} B}{\sin Acos A - \sin Bcos B} = \frac{2 \sin (A + B) \sin (A - B)}{\sin 2 A - \sin 2 B} \\ = \frac{2 \sin (A + B) \sin (A - B)}{2 \sin (A - B) \cos (A + B)} \\ = \tan (A + B) \end{array}$

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